Find the value of a, if the line passing through (–5, –8) and (3, 0) is parallel to the line passing through (6, 3) and (4, a). -Maths 9th

Find the value of a, if the line passing through (–5, –8) and (3, 0) is parallel to the line passing through (6, 3) and (4, a). -Maths 9th
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1 Answer

Answer :

Let C(x, y) be the centre of the circle passing through the points P(6, –6), Q(3, –7) and R(3, 3) Then, PC = QC = RC(Being radius of the same circle) PC2 = QC2 ⇒ (x – 6)2 + (y + 6)2 = (x – 3)2 + (y + 7)2 ⇒ x2 – 12x + 36 + y2 + 12y + 36 = x2 – 6x + 9 + y2 + 14y + 49 ⇒ –12x + 12y + 6x – 14y + 72 – 58 = 0 ⇒ – 6x – 2y + 14 = 0 ⇒ 3x + y –7 = 0          ...(i) Also, QC2 = RC2 ⇒ (x – 3)2 + (y + 7)2 = (x – 3)2 + (y – 3)2 ⇒ x2 – 6x + 9 + y2 + 14y + 49 = x2 – 6x + 9 + y2 – 6y + 9 ⇒ 14y + 6y = 9 – 49 ⇒ 20y = – 40 ⇒ y = –2             ...(ii) Putting y = –2 in (i), we get 3\(x\) + (–2) – 7 = 0 ⇒ 3\(x\) = 9 ⇒ \(x\) = 3 ∴ The centre is (3, –2).
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A(5,0) and B(0,8) are two vertices of triangle OAB. a). What is the equation of the bisector of angle OAB. b). If E is the point of intersection of this bisector and the line through A and B,find the coordinates of E. Hence show that OA:OB = AE:EB -Maths 9th

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Last Answer : This answer was deleted by our moderators...

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Last Answer : Its a simple question which have a simple answer.. The answer is that :- When you draw a straight line on a paper you will find that the line is straight and when you measure it by you protector or ... so that means 90°+90°=180° which means it will never interest each other at any condition...

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Draw the graph of the equation represented by a straight Line which is parallel to the X-axis and at a distance 3 units below it. -Maths 9th

Description : Draw the graph of the equation represented by a straight Line which is parallel to the X-axis and at a distance 3 units below it. -Maths 9th

Last Answer : Any straight line parallel to X-axis in negative direction of Y-axis is given by y = - k, where k is the distance of the line from the X-axis. Here, k = 3. Therefore, the equation of the line is y = -3. To ... , plot the points (1,-3), (2, -3) and (3, -3) and join them. This is the required graph.

Prove that the line joining the mid-points of the diagonals of a trapezium is parallel to the parallel sides of the trapezium. -Maths 9th

Description : Prove that the line joining the mid-points of the diagonals of a trapezium is parallel to the parallel sides of the trapezium. -Maths 9th

Last Answer : Given Let ABCD be a trapezium in which AB|| DC and let M and N be the mid-points of the diagonals AC and BD, respectively.

If a line is drawn parallel to the base of an isosceles triangle to intersect its equal sides, prove that the quadrilateral, so formed is cyclic. -Maths 9th

Description : If a line is drawn parallel to the base of an isosceles triangle to intersect its equal sides, prove that the quadrilateral, so formed is cyclic. -Maths 9th

Last Answer : Given ΔABC is an isosceles triangle such that AB = AC and also DE || SC. To prove Quadrilateral BCDE is a cyclic quadrilateral. Construction Draw a circle passes through the points B, C, D and E.

Why parallel line can not intersect -Maths 9th

Description : Why parallel line can not intersect -Maths 9th

Last Answer : Its a simple question which have a simple answer.. The answer is that :- When you draw a straight line on a paper you will find that the line is straight and when you measure it by you protector or ... so that means 90°+90°=180° which means it will never interest each other at any condition...